The point of intersection C of the plane 8x + y + 2z = 0 and the line joining the points A(-3, -6, 1

Q: Why is the line equation $\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$ written in such a weird way?

It is a classic JEE trap! The standard form of a line is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. The term $1-x$ has a negative coefficient for $x$. You must rewrite it as $\frac{x-1}{-1}$ to correctly identify the direction vector.

Q: Why do we use a parameter $\mu$ for point $D$?

Because $D$ is not a fixed point; it is the foot of the perpendicular. By using $\mu$, we define the entire line as a set of points. The specific value of $\mu$ that satisfies the perpendicularity condition gives us the exact location of $D$.

Q: How do I know if my direction ratios are correct?

Direction ratios are not unique; they are proportional. If you get $(-5/42, -130/42, 85/42)$, you can multiply by a constant (like $42/5$) to get integers. The problem asks for coprime integers, so ensure you divide by the greatest common divisor.

Question

The point of intersection $C$ of the plane $8x + y + 2z = 0$ and the line joining the points $A(-3, -6, 1)$ and $B(2, 4, -3)$ divides the line segment $AB$ internally in the ratio $k : 1$. If $a, b, c$ ($|a|, |b|, |c|$ are coprime) are the direction ratios of the perpendicular from the point $C$ on the line $\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$, then $|a + b + c|$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Intersection of Line and Plane

Imagine a plane defined by the equation $8x + y + 2z = 0$. A line segment connects points $A(-3, -6, 1)$ and $B(2, 4, -3)$.

The point $C$ where this line pierces the plane divides the segment $AB$ in a ratio $k:1$. Using the section formula, the coordinates of $C$ are:

$$x = \frac{2k-3}{k+1}, \quad y = \frac{4k-6}{k+1}, \quad z = \frac{-3k+1}{k+1}$$

Since point $C$ lies on the plane, it must satisfy the plane's equation. Substituting these coordinates into $8x + y + 2z = 0$ gives:

$$8\left(\frac{2k-3}{k+1}\right) + \left(\frac{4k-6}{k+1}\right) + 2\left(\frac{-3k+1}{k+1}\right) = 0$$

Clearing the denominator $(k+1)$, we obtain the linear equation:

$$8(2k-3) + (4k-6) + 2(-3k+1) = 0$$

Simplifying this expression leads to $14k - 28 = 0$, which reveals that $k=2$. Substituting $k=2$ back into our coordinate expressions, we find the point $C$ to be:

$$C = \left(\frac{1}{3}, \frac{2}{3}, -\frac{5}{3}\right)$$

### The Perpendicular Challenge

We now consider the line $L$ given by $\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$. To identify the direction vector, we rewrite the equation in standard symmetric form:

$$\frac{x-1}{-1} = \frac{y+4}{2} = \frac{z+2}{3}$$

The direction vector of line $L$ is $\vec{v} = (-1, 2, 3)$. Let $D$ be the foot of the perpendicular from $C$ to line $L$. Any point $D$ on $L$ can be expressed using a parameter $\mu$:

$$D = (-\mu + 1, 2\mu - 4, 3\mu - 2)$$

The vector $\vec{CD}$ is calculated as $D - C$:

$$\vec{CD} = \left(-\mu + 1 - \frac{1}{3}, 2\mu - 4 - \frac{2}{3}, 3\mu - 2 + \frac{5}{3}\right) = \left(-\mu + \frac{2}{3}, 2\mu - \frac{14}{3}, 3\mu - \frac{1}{3}\right)$$

### The Final Synthesis

For $CD$ to be perpendicular to line $L$, the dot product $\vec{CD} \cdot \vec{v}$ must equal zero:

$$-1\left(-\mu + \frac{2}{3}\right) + 2\left(2\mu - \frac{14}{3}\right) + 3\left(3\mu - \frac{1}{3}\right) = 0$$

Solving this equation yields $\mu = \frac{11}{14}$. Substituting $\mu$ back into the expression for $\vec{CD}$, we obtain the direction ratios:

$$\vec{CD} = \left(-\frac{5}{42}, -\frac{130}{42}, \frac{85}{42}\right)$$

To find the coprime integers $a, b, c$, we scale the vector by multiplying by $\frac{42}{5}$, resulting in the vector $(-1, -26, 17)$. The final required value is:

$$|a + b + c| = |-1 - 26 + 17| = |-10| = \mathbf{10}$$

Enter Numerical Value:

Visualized Solution

Visualizing the Intersection

Applying the Section Formula

Satisfying the Plane Equation

Solving for $k$

Finding Coordinates of $C$

Analyzing the Second Line

General Point on the Line

Direction Ratios of $C D$

Perpendicularity Condition

Solving for $μ$

Final Direction Ratios

The Final Answer

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Intersection of Line and Plane

The Perpendicular Challenge

The Final Synthesis

Similar Questions

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